Power Series
Last updated
Last updated
Try to think
Power series = Geometric Series.
►Refer to Math24: Power series
Power series is actually the Geometric series in a more general and abstract form.
For easier to remember it, that could be simplified as:
In this function it's critical to know that: a_n IS A CONSTANT NUMBER! NOT A VARIABLE !
Differentiate Power series►Refer to Khan academy: Differentiating power series
We have 2 ways to differentiate series, they work same way:
One way is to expand the series with real numbers (1,2,3...) and take their derivatives:
(▲Note that this is constantly true for power series.)
Another way is to calculate the term's derivative and then plug in the real numbers (1,2,3):
Either way will do, it depends on the actual equation for you to choose which way you're gonna use.
Solve:
Let's organize this function to make it clear:
Solve:
Notice that: If we plug in the x=0 at beginning, everything will be 0 and we don't have anything to calculate.
So Let's keep the x in the terms until the last step.
It's easier to expand the series with real numbers, and we're to try 3 or 4 terms in this case.
Because at the end of it you'll notice, if we're doing Third Derivative, then more than 3 terms will just bring more 0s.
First we're to organize the function in the standard power series form:
And the constant number a_n in the function is:
Let's plug in the real number (0,1,2,3,4) for n to expand the series:
Based on that we can take the derivatives:
Now we've got the third derivative, so let's plug in the x=0 and see what we get:
And that's the answer.
And now you know why in this case it's a waste to calculate more terms.
Integrate Power SeriesSolve:
Knowing that Integrate a series can be turned to a series of Integrations of terms:
Integrate the terms:
And we get a simple Geometric series. So that we can apply the formula of calculating geometric series:
Let's apply the formula:
► Jump over to have practice at Khan academy.
Solve:
Let's assume the function of the series above is f(x), and the series below is g(x)
It's clear to see that the g(x) is theAntiderivativeof thef(x)`.
So we just need to integrate the function of the f(x):
We see that the antiderivative can represent the g(x), but only with the C in it:
So we need to solve for C. The easiest way is to plug in 0 for g(x):
Then the answer is:
Solve:
Set the two series as f(x) and g(x).
It's clear that g(x) = -f'(x).
So by differentiate f(x) we will get:
Then -f'(x) would be:
